Seely F.B. Analytical Mechanics for Engineers
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ALGEBRAIC
METHOD
41
principle
of
moments,
is
equal
to the
algebraic
sum of the moments
of
the
forces
of
the
system.
Hence,
if the resultant of
a
coplanar
parallel
force
system
is
a
couple
the
moment, C,
of the
couple may
be
determined
from the
expression,
the
aspect
of
the
couple,
of
course,
being
the same
as
that of
the
plane
of the
forces.
If both
2F
and
1>M
equal
zero the resultant
is
equal
to zero.
ILLUSTRATIVE
PROBLEMS
26.
Fig.
38
represents
a beam
resting
on
two
supports
and
carrying
four
loads
as
indicated.
Find
the
resultant
of
the
loads.
2000
lb.
1500 lb.
42
RESULTANTS
OF
FORCE
SYSTEMS
27.
An
arm
mounted on an
axle
at
(Fig.
39)
is acted
on
by
the four
forces
shown.
Determine
the resultant of
the forces.
20 Ib.
24
lb.
15 lb.
24
Ib.
forces,
figure.
Solution.
R
=
ZF
=
15
+
15 -20
-10=0.
Therefore the
resultant is
not
a
force,
but is
a
couple, C,
the
moment
of
which
may
be found
by
taking
mo-
ments
about 0.
Thus,
C
=
SM
=
20X10
-15X6 -10X5
+15
X
12
=
+240
lb
.-in.
Hence
any couple
having
a
moment
of
+240
Ib.-in. will
produce
the
same
external effect on the
arm
(change
in
the
rotational
motion)
as
the
four
given
One
resultant
couple
having
forces
equal
to 24 lb. is shown
in the
15
lb.
FIG.
39.
PROBLEMS
28.
Fig.
40
represents
a beam
12
ft.
long
acted
upon by
five forces
as
shown
.
Find
the anti-resultant
of the
five
forces.
29. Five forces
act
on
a
body
as
shown
in
Fig.
41.
Replace
200
lb.
the two
60-lb.
forces
by
a
single
150
n,.
force and
then
find the resultant
of
this
force
and
the
remaining
three
forces.
30.
A force of
10
lb. acts
along
the
2/-axis,
the sense of
the force
being
positive.
Resolve
the
force
into
components
P
and
Q
along
the lines
x
=
l
and
#
=
3, respect-
ively.
Ans.
P
=
151b.
Q=-51b.
31.
A
bar
6 ft.
long
is
hinged
at
A and
is acted
on
by
five
forces as shown
in
Fig.
42. Determine the
resultant
of the
five forces.
Ans.
C=+901b.-ft.
32. A bar
10 ft.
long
is acted
on
by
the forces shown
in
Fig.
43.
Find
the
equilibrant
of
the force
system
and
represent
it
acting
on
the
bar.
Ans.
Equilibrant
=
+15
lb.,
2f
ft. to
the left of
the
20-lb.
force.
33.
The
40-lb. force
shown
in
Fig.
44 is the resultant of
the
10-lb.
force
and
a force P
not
shown.
Determine
P
completely
by
means
of
a
force
and
a
funicular
polygon
and
check
the result
by
the
algebraic
method.
ALGEBRAIC
METHOD
43
In each
of the
following
three
problems
find the
resultant
of
the
force
system.
The forces in
each
problem
are
parallel
to
the
y-axis.
The
values
80
Ib.
44 RESULTANTS OF FORCE
SYSTEMS
27.
Couples.
In
Art.
15 a
couple
is
defined
as
two
equal,
parallel,
non-collinear
forces of
opposite
sense. The
moment of
a
couple
is defined as the
algebraic
sum
of
the
moments of
the two
forces of the
couple,
and it is
shown
that the
moment is the same
for
all
moment-centers
in
the
plane
of the
couple.
Further,
it is
shown
that the characteristics of a
couple
(moment, aspect,
and
sense)
may
be
completely represented
by
means of a vector.
Since
couples
play
an
important
part
in
engineering problems
it
will be of
advantage
to
consider
certain additional
propositions
and
transformations
by
means
of which
couples
can
easily
be
combined with other
couples
or with
other forces.
PROPOSITION
I. A
couple may
be
translated into
any parallel
position
in its
own
plane
or
into
any parallel
plane
without
changing
its
&xternal
effect
on the
body
on
which it
acts.
Proof.
Consider
the
couple Pp having
a
moment-arm, AB,
equal
to
p
as
shown
in
Fig.
45.
Construct
a
parallelogram
having
AB
as
one side
and
CD as
p
^"
the
opposite
side.
This
par-
allelogram may
or
may
not
be in
the
plane
of
the
given
couple.
At
each
of
the
B
points
C
and D introduce
two forces
of
opposite
sense
which -are
parallel
to the
forces of the
given
couple
and
P
equal
to them
in
magnitude.
The force
P at A
and the
upward
force
P
at
D
may
be
replaced
by
their
resultant.
This resultant
is
an
upward
force, 2P,
the
action
line
of
which is
parallel
to the
ac-
tion
lines of the
forces
of
the
given couple
and
is locat-
ed
half
way
between
them.
Hence
the
action
line
passes
through 0,
the
point
of
intersection
of
the
diagonals
of
the
parallelogram.
Similarly,
the
force
P at
B
and the
downward
force
P
at
C
may
be
replaced
by
a downward
force, 2P, acting through parallel
to
the
forces
of the
given
couple.
The two forces
2P,
2P
at
cancel,
thereby
leaving
only
COUPLES
45
the
upward
force
P
at C
and
the
downward
force P
at D.
These
two
forces constitute
a
couple
which is
equivalent
to the
given
couple.
Hence the
proposition
is
established.
PROPOSITION
II.
A
couple
may
be
rotated
through
any angle
in
its
plane
without
changing
its external
effect
on
the
body
on
which
ii
acts.
Proof.
In
Fig.
46 the
two
equal
forces,
PI
and
P^
consti-
tute a
couple
having
a
moment-arm, AB,
equal
to
p.
Through
0,
the
midpoint
of the
arm,
draw
any
line
CD
making
an
angle
6
with
AB
and
scale off OD and OC
each
equal
to
%p.
At
C in-
troduce two
opposite
forces,
PS
and
P4,
which are
perpen-
dicular to
CD,
each of
which
is
equal
in
magnitude
to the
forces
PI
and
PZ
of the
given
couple. Similarly
at
D
intro-
duce
two
forces,
PS
and
PQ,
perpendicular
to
CD,
each
being
equal
in
magnitude
to
the forces of the
given couple.
Prolong
the action line of
PI
until it
intersects
the action
line of
P4
at E. The resultant
of the
two
forces
PI
and
P
is a force the
action
line of which
bisects the
angle
AEC
and
hence
coincides
with
EO. Let R denote
the
magnitude
of
this
force.
In the same
way
the resultant of
P*>
and
P<z
may
be
shown
to be
a
force
of
magnitude
R
the action
line
of which is
FO.
Since
these
two forces
R,
R
are
equal,
collinear,
and
opposite
they cancel,
thereby
leaving
the forces
PS
and
PQ
which
constitute a
couple
having
a
moment-arm CD
which,
by
construction,
is
equal
to
p.
This
equivalent
couple,
therefore,
is
merely
the
given
couple
rotated
through
an
angle
6.
PROPOSITION
III. The
magnitudes
of
the
forces
of
a
couple
and
the moment-arm
of
the
couple
may
be
changed
to
any
values
without
changing
the
external
effect of
the
couple,
provided
that
the
product
of
either
force
and the moment-arm
remains
constant.
Proof.
Given
the
couple
Pp
(Fig.
47)
having
a
moment-arm,
AB,
equal
to
p.
Let the
arm
be
increased to
CD
=
(p-}-2x)
as
FIG.
46.
46 RESULTANTS
OF FORCE SYSTEMS
shown.
At C introduce two forces of
opposite
sense each
equal
to
Pp
p+2x
and
perpendicular
to
CD.
Similarly,
introduce
two forces
Pv
Pv
.
n
at
D. The
force
P
at
A
and
the
upward
force
~-
at
D
p+2x
p+2x
may
be
replaced by
their
resultant,
which
is an
upward
force of
magnitude
~=
,
the direction of the line of
action
being per-
pendicular
to
CD.
The
position
of the line of action of
this
result-
COUPLES
47
-p.
This
proposition
can
be
proved
by
means of
the
three
trans--
formations
discussed above.
It is
sufficient
to
prove
the
propo-
sition
for
two
couples,
for,
if
two
couples
can be
combined
into
a
single
resultant
couple, evidently
this
couple
can be combined
with
a
third
coHiple,
and so on.
Let
the two
couples
Pp
and
Qq (Fig.
48a)
act on
a
body.
The
magnitude
of the
forces
of the second
couple may
be
changed
from
Q
to
P
by changing
the moment-
arm
from
q
to
^
(Fig.
486),
as
discussed
in
Proposition
III.
The second
couple
can
then
be rotated
until
the
forces
of
the two
couples
are
parallel
(Fig.
48c)
. The second
couple
can now
be
translated
until one
of its forces
is
collinear with
one of the
forces of the
first
couple
(Fig.
48d).
This
trans-
lation
can
always
be
made
in such a
way
that
these two
equal,
collinear
forces
are
opposite
in
sense
and
hence
will
can-
cel, thereby
leaving
the
two
equal,
parallel,
and
opposite
forces
P,
Pwhich
constitute
a
couple.
The
arm of
this resultant
couple
is
(P+
p )
'
moment
is P
(c)
(*)
FIG.
48.
-M }
=
Pp
+
Qq.
Hence the
mo-
ment of
the resultant
is
equal
to the
sum
of
the
moments
of
the
given couples
as
stated
in the
proposition.
PROPOSITION
V.
Two
coplanar
couples
which
have
the
same
moment and sense are
equivalent.
This
proposition
follows
from
the
three
transformations
pre-
48
RESULTANTS
OF
FORCE
SYSTEMS
viously discussed, since, by
means
of
them,
one
of
the
couples
may
be
superimposed
on
the
other.
4.
NON-CONCURRENT,
NON-PARALLEL FORCES IN A
PLANE
28.
Graphical
Methods. First
Method. The
resultant of
a
system
of
non-concurrent,
non-parallel
forces
in
a
plane
is
either
a
force
or
a
couple.
If the
resultant is a force it
may
be
determined
by
use of the
parallelogram
law. In
Fig.
49
are
shown
three
non-concurrent, non-parallel
forces
FI,
F2,
and
F%
acting
on a
body.
The
forces
F\
and
F^
may
be
combined into
a
single
resultant
force
R\. R\
and
Fa
may
be combined into
a
FIG.
49.
'
single
force
R%
which is
the resultant
of the
given
system
of
forces.
If the resultant
obtained
by
combining
all,
except one,
of
the forces of
such a
system
is
equal
to that
one and
is
parallel
to
it and of
opposite
sense,
the
two
forces
constitute
a
couple.
Furthermore,
if the
action
lines
of
these two
forces
are
collinear
the
forces cancel and the
resultant
of
the
system
is
equal
to
zero.
Second Method.
The resultant of
a
sytem
of
non-concurrent,
non-parallel
forces
in a
plane
may
also
be
found
by
the con-
struction
of
a force
and a
funicular
polygon.
This
method
is
of
great
importance
in
engineering problems.
The
method
of
con-
struction of these
polygons
is the
same as
that
described
in
con-
THE PRINCIPLE OF MOMENTS
49
nection with a
system
of
parallel
forces
(Art. 24)
and will not
here-
be discussed
in detail
for
the
non-parallel
force
system.
The
method will
be
illustrated, however,
by
the
following problem.
ILLUSTRATIVE
PROBLEM
37.
A
beam
9
ft.
long
is
acted
on
by
four forces as shown in
Fig.
50.
Determine
the resultant of
the
forces
by
use of
the
force and
funicular
polygons.
Solution.
The force
polygon
as shown in
Fig. 50(6)
is constructed
by
laying
off
the vectors
AB,
BC,
CD,
and DE which
represent
the
magni-
tudes and the directions
of the
given
forces.
The
closing
side AE
of
the
polygon
represents
the
resultant force
in
magnitude
and
in
direction.
By
measuring
AE
to scale
the
magnitude
of
the resultant force is found
Scale.
1
in.
=
8
ft.
(a)
Scale.
1
in.=
20001b.
(6)
FIG. 50.
to be
2450 Ib. and
the line AE is found to
make an
angle
of
80 with
the
horizontal.
In
Fig.
50
(a)
is shown
the
funicular
polygon
in
which one
point
on
the
action line of the resultant is
found,
namely,
the
intersection
of
oa
and
oe.
Therefore
the
action line
of the
resultant
passes
through
this
point
and
is
parallel
to
AE.
By
measurement the
action
line is
found to
intersect the
beam at a
point
3.6
ft. from the
left end
of the
beam.
Hence,
if the
four forces were
replaced by
a
single
force, R,
of 2450 Ib.
as shown
in
Fig.
50(a),
the
reactions
at
the
ends of
the
beam
(external
effects)
would be
unchanged.
29.
Principle
of
Moments.
The
algebraic
sum
of
the
moments
of
the
forces of a
coplanar,
non-concurrent,
non-parallel
force
system
about
any point
in
the
plane
of
the forces
is
equal
to
the
50 RESULTANTS
OF
FORCE
SYSTEMS
moment of the
resultant
of the
system
about the same
point.
The
proof
is identical with
that
given
for a
system
of
parallel
forces
(Art. 25)
and the
principle
will be
used,
in
the
algebraic
method,
for
finding
the
action
line of the resultant force or the
moment of the
resultant
couple
similar to the
way
it was used for
parallel
forces
(Probs.
26 and
27).
30.
Algebraic
Method.
The resultant of a
system
of
non-con-
current, non-parallel
forces in a
plane
is either a force or a
couple
as
explained
in
Art.
28.
If the resultant is a force
it
may
be
deter-
mined
as
follows: In
Fig.
51
(a)
is
represented
a
body
acted on
by
four
forces F
f
,
F",
F'",
and
F"".
Let each
force
be
replaced
7
_'/
A.
4
7
F'
FIG. 51.
by
its x- and
^/-components.
The
original
force
system
is now
resolved
into two
systems
of
parallel
forces,
the resultant
of
each
parallel S3
r
stem
being
a
force
parallel
to
the forces
of
the
system.
According
to
Art. 26 the resultant of
the
com-
ponents
in the z-direction
is 2F
X
and
similarly,
the resultant
of
the
components
in the
^/-direction
is
2Fj/.
Hence,
the
magnitude
and
the direction' of the resultant
of
these two forces
(which
is
also
the resultant of the
original forces)
may
be
found
from
the
equations
R
=
V
(ZF
X
)
2
+(2FV)
2
and tan6
x
=
(Fig.
516),
in which
6
X
is the
angle
which
the
action
line
of
R
makes
with
the
#-axis.
In
order
to
obtain
the
position
of the
line
of action
of
the resultant
force
R
the
principle
of
moments
may
be